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Is $\mathbb{R}\setminus\{-2,6\}$ a neighbourhood of $0$? When we consider the $\mathbb{R}$ with the Euclidean topology (i.e. the topology induced by the euclidean metric).

I wrote that it is since:

$0 \in \mathbb{R}\setminus\{-2,6\} \subseteq \mathbb{R}\setminus\{-2,6\}$ and $\mathbb{R}\setminus\{-2,6\}$ is an open set in $\mathbb{R}$ with the Euclidean topology (since it is an open set in $\mathbb{R}$ with the Euclidean metric.

Is this correct???

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    Yes this is perfectly correct ! – 2017-01-16
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    What is your definition of neighbourhood? Different authors use different notions. – 2017-01-16
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    I don't really get why you write "$\mathbb{R} - \{-2,6\} \subseteq \mathbb{R} - \{-2,6\}$. Did you mean $\mathbb{R} - \{-2,6\} \subseteq \mathbb{R}$? – 2017-01-16
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    @Brahadeesh : do you have a definition of neighborhood such that $\mathbb R - \{-2,6\}$ is not a neighborhood of $0$? – 2017-01-16
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    @N.H. Fair point, I realised that only after I commented. – 2017-01-16
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    T102 : I wrote that this is perfectly correct but as SteamyRoot pointed you don't need to say $\mathbb R - \{-2, 6\} \subset \mathbb R - \{-2, 6\}$. $x \in U$ and $U$ is open in $\mathbb R$ is a sufficient condition for be a neighborhood. – 2017-01-16
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    You should also mention the fact that $0\in\mathbb R\setminus\{-2,6\}.$ For instance, $\mathbb R\setminus\{-2,0,6\}$ is also an open set in the Euclidean topology, but it's *not* a neighborhood of $0.$ – 2017-01-16

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To elaborate a bit on some of the comments above.

Note that $\mathbb{R} - \{-2, 6\} = (-\infty, -2) \cup (-2, 6) \cup(6, \infty)$.

Since we have $\left\{ (-\infty, -2), (-2, 6),(6, \infty)\right\} \subset \mathcal{T}$, where $\mathcal{T}$ is the standard topology on $\mathbb{R}$, we can see that $\mathbb{R} - \{-2, 6\} \in \mathcal{T}$, and thus $\mathbb{R} - \{-2, 6\}$ is an open set containing $0$ and is, by definition, a neighbourhood.